Problem: Multiply the following complex numbers: $({2i}) \cdot ({-2+4i})$
Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({2i}) \cdot ({-2+4i}) = $ $ ({0} \cdot {-2}) + ({0} \cdot {4}i) + ({2}i \cdot {-2}) + ({2}i \cdot {4}i) $ Then simplify the terms: $ (0) + (0i) + (-4i) + (8 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 0 + (0 - 4)i + 8i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 0 + (0 - 4)i - 8 $ The result is simplified: $ (0 - 8) + (-4i) = -8-4i $